Is there another mathematical way to approach this problem? 
$$f(x)-f(x-1) = x-5$$
$$f(16) = 74$$
Compute $f(1)$.

So, this is basically a linear function. I, however, calculated it without taking the easier way to approach this problem.
If
$$f(16) - f(15) = 11$$
Then
$$74-f(15) = 11 \implies f(15)=63$$
This will take so long as it seems. Since we're given a linear function, how would we compute it in an other mathematical way? Furthermore, is there a general rule for what you advise?
 A: Hint: Notice that you can write $f(1)$ as
\begin{align}
f(1) &= f(1)-f(2)+f(2) -f(3)+\ldots-f(16)+f(16)\\
&= \sum_{k=2}^{16} (f(k-1)-f(k)) + f(16)\\
&= \sum_{k=2}^{16} (5-k) + 74.
\end{align}
A: First, note that $f$ isn't linear at all (if it were linear, then $f(x)-f(x-1)$ would be constant. However, that difference does suggest something quadratic: its second difference will be constant. Thus, we'll try to find $a$, $b$, and $c$ such that $f(x) = ax^2 + bx + c$ satisfies this. For this purpose, we garner some equations relating $a$, $b$, and $c$: 
First, $f(16) = 74$ tells us that $256a + 16b + c = 74$. Second, $f(x)-f(x-1) = x - 1$ gives us $a(x^2 - (x-1)^2) + b(x - (x-1)) = x - 1$. Simplifying that gives us $a(2x-1)+b=x-1$, for all $x$. Taking $x = 0$ gives us $b - a = -1$, $x = 1$ give $a + b = 0$, so $b = -a$, and the $x = 0$ case gives $2a = 1$, so $a = \frac{1}{2}$, and $b = \frac{-1}{2}$. Finally, $f(16)=74$ gives us $120 + c = 74$, so $c = -46$. Thus, $f(x) = \frac{1}{2}x^2 - \frac{1}{2}x - 46$. 
You can quickly go back and check that this satisfies all relevant conditions, if you like (it does), and therefore $f(1) = \frac{1}{2}(1^2) - \frac{1}{2}(1)-46 = -46$. 
A: Hint:
Let $f(x)=g(x)+a+bx+cx^2$
$$x-5=f(x)-f(x-1)=g(x)-g(x-1)+c(2x-1)+b$$
Set $2c=1,b-c=-5$ so that $$T=g(x)=g(x-1)$$
For any integer $x$
$$T=\cdots=g(16)=f(16)-a-b(16)-c(16^2)$$ 
A: $$f(16)-f(15) = 16-5$$
$$f(15)-f(14) = 15-5$$
$$f(14)-f(13) = 14-5$$
$$ \vdots$$
$$f(3)-f(2) = 3-5$$
$$f(2)-f(1) = 2-5$$
So, if we sum this we get  $$f(16)-f(1) = (16+15+...+3+2)-15\cdot 5$$
and thus $$f(1) = 75-{15\cdot 17\over 2}+f(16)$$
A: Hint.
Try 
$$
f(x) = \frac 12(x-4)(x-5) + C_0
$$
The linear recurrence equation can be solved as
$$
f(x) = f_h(x)+ f_p(x)\\
f_h(x) = C_1\\
f_p(x) = \frac 12(x-4)(x-5)+C_2
$$
with
$$
f(16)= 74 \to C_0 = 8
$$
